Problem: In triangle $ABC$, $\cos(2A-B)+\sin(A+B)=2$ and $AB=4$. What is $BC$?
Answer: The only way that the sum of a cosine and a sine can equal 2 is if each is equal to 1, so
\[\cos (2A - B) = \sin (A + B) = 1.\]Since $A + B = 180^\circ,$ $0 < A + B < 180^\circ.$  Then we must have
\[A + B = 90^\circ.\]This means $A < 90^\circ$ and $B < 90^\circ,$ so $2A - B < 180^\circ$ and $2A - B > -90^\circ.$  Hence,
\[2A - B = 0^\circ.\]Solving the equations $A + B = 90^\circ$ and $2A = B,$ we find $A = 30^\circ$ and $B = 60^\circ.$

[asy]
unitsize(1 cm);

pair A, B, C;

A = 4*dir(60);
B = (0,0);
C = (2,0);

draw(A--B--C--cycle);
draw(rightanglemark(A,C,B,10));

label("$A$", A, N);
label("$B$", B, SW);
label("$C$", C, SE);
label("$4$", (A + B)/2, NW);
[/asy]

Therefore, triangle $ABC$ is a $30^\circ$-$60^\circ$-$90^\circ$ triangle, so $BC = \frac{AB}{2} = \boxed{2}.$